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swat32
3 years ago
14

Please help :( A ray of laser light strikes a glass surface at an angle θa = 22.5° to the normal. What is the angle θb of the re

fracted ray? Let the index of refraction for glass nb = 1.77 (you are able to choose two)
A 12.5°
B 12.7°
C 10.1°
D 31.8°
Physics
1 answer:
pshichka [43]3 years ago
4 0

Answer:

Explanation:

We shall apply law of refraction which is as follows

sin i / sinr = μ  , where i is angle of incidence , r is angle of refraction and μ is refractive index

here i = θa = 22.5°

r = θb

μ  = 1.77

sin22.5 / sinθb = 1.77

.3826 / sinθb = 1.77

sinθb = .216

θb  = 12.5 °.

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Full solution calculation can be found in the attachment below.

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Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
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Answer:

a

\lambda = 3.68 *10^{-36} \  m

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\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

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So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

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Generally the energy of the proton is mathematically represented as

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=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

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so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

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